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Let $\alpha_n \in \mathbb C$ and $\lim_{n\to\infty}\alpha_n = 0$. Let $T$ be a linear continuous operator from $\ell^p \to \ell^p (1\le p\le \infty)$ defined by $$ T((x_1, x_2, \ldots)) = (\alpha_1 x_1, \alpha_2 x_2 , \ldots). $$ Then I want to show that $T$ is compact operator on $\ell^p$, which means any bounded sequence in $\ell^p$, there exists a subsequence $\{x_{n_k}\}$ such that $\{Tx_{n_k}\}$ converges in the sense of norm.

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Let $$(T_n(\underline x))_k:=\begin{cases}\alpha_kx_k&\text{if }k\le n\\ 0&\text{if }k>n\end{cases}$$

You can prove that $$\left\lVert T-T_n\right\rVert_{\mathfrak L(\ell^p,\ell^p)}\le \sup\{\lvert \alpha_k\rvert\,:\,k>n\}\stackrel{n\to\infty}\longrightarrow 0$$

So $T$ is limit in $\mathfrak L(\ell^p,\ell^p)$ of finite-rank operators. Hence, it is compact.

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  • $\begingroup$ Really appreciate to you. Now I got it $\endgroup$ – John S Jul 23 '16 at 17:20

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